Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**johny****Member**- Registered: 2005-11-19
- Posts: 34

Hey, i need some help on the shape.....its basically a truncated cone, and i need to take 2 equation i.e the surface area formula being the one and the volume formula the other (volume is fixed to 600cm3). So i need to substitue volume equation in the surface area, by re-arraning the height (h) in the volume, and putting in the h of surface area, if u get what i mean??? I havent got anyother values, only the volume, that being 600cm3.....please help guys.

Offline

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

A truncated cone is basically a normal cone with the top cut off. That means that to find the volume, we need the radius of the top circle, the radius of the bottom circle and the height. If we only know one the height and the volume, then we cannot find the other two because it is two variables but only one equation.

Why did the vector cross the road?

It wanted to be normal.

Offline

**johny****Member**- Registered: 2005-11-19
- Posts: 34

Hmmm well i didnt want the volume or area......i just wanted an equation. by substituting the height in volume and putting that equation of height in the formula of Surface area for the truncated cone??? Sorry if i am confusing u!!!!

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,529

The volume of a truncated cone would be the volume of an **un**truncated cone minus the volume of the bit chopped off.

The volume of a normal Cone = π × r² × (h/3)

If the cone were chopped off at "y" then we would remove a smaller cone, whose height would be (h-y) and whose base radius would be r × (h-y)/h

Hence the "removed cone"s volume would = π × (r × (h-y)/h)² × ((h-y)/3)

So the Volume of the truncated cone = π × r² × (h/3) - π × (r × (h-y)/h)² × ((h-y)/3)

The areas of a normal cone are:

Base = π × r²

Surface Area of Side = π × r × s (where s = side length)

The areas of a truncated cone are (h=height of normal cone, y=height of truncated cone):

Base = π × r²

Top = π × (r × (h-y)/h)²

Surface Area of Side = π × r × s × (y/h)

That's a start, but I gotta go ...

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

Offline

**johny****Member**- Registered: 2005-11-19
- Posts: 34

Thanks man, really helpful .....will wait for ur other half.

Offline

Pages: **1**